How do I make a different shape like a hexagon or triangle using image processing?

Hi, I have a Labtest in 2 hours and just found out that we need to know how to make different shapes such as a hexagon using image processing which I’m not familiar with. We were given this code which just makes a square. Any help would be appreciated.

layer_1 = 255*ones(50, 50);

layer_2 = 255*ones(50, 50);

layer_3 = 255*ones(50, 50);

my_image(:,:,1) = layer_1; % applied to 3D matrix. my_image(:,:,2) = layer_2; % applied to 3D matrix. my_image(:,:,3) = layer_3;

imshow(my_image);

for row_index = 1:25 for col_index = 25:50 my_image(row_index,col_index,1)= 0; my_image(row_index,col_index,3) = 0; end end

imshow(my_image);

My friend also gave me this code but it also makes a square. for i=1:100 for j=1:100 mat(j,i,1)=255; mat(j,i,2)=0; mat(j,i,3)=0; end end

imshow(mat)

Also pls bare my shitty editing skills, I’m not sure how to insert code since I’m using my iPad rn.

Thanks

if true
  for i=1:100
  for j=1:100
mat(j,i,1)=255;
mat(j,i,2)=0;
mat(j,i,3)=0;
  end
end
imshow(mat)
end

if true
  layer_1 = 255*ones(50, 50);

layer_2 = 255*ones(50, 50);

layer_3 = 255*ones(50, 50);

my_image(:,:,1) = layer_1; % applied to 3D matrix. my_image(:,:,2) = layer_2; % applied to 3D matrix. my_image(:,:,3) = layer_3;

imshow(my_image);

for row_index = 1:25 for col_index = 25:50 my_image(row_index,col_index,1)= 0; my_image(row_index,col_index,3) = 0; end end

imshow(my_image); end

Answer 1

[Akira Agata]

If you want to create regulay polygon shape image, how about using nsidedpoly function?

The folloing is an example:

% Create hexagon polyshape
pgon = nsidedpoly(6,'Center',[50,50],'Radius',30);

% Generate 100x100 binary image with hexagon
[xGrid,yGrid] = meshgrid(1:100,1:100);
BW = isinterior(pgon,xGrid(:),yGrid(:));
BW = reshape(BW,size(xGrid));

% Visualize
figure
imshow(BW)

[HG]

https://www.mathworks.com/matlabcentral/answers/846710-display-geometric-shapes-and-output-matrices?s_tid=srchtitle

how can I create a triangle or rectangle with curve edge ? and then show the matrix with imagesc ?

Answer 2

[Image Analyst]

See my attached demos.

[Sakibur Rahman]

Sorry but is there a simpler code or explanation? I just want to know how to create shape like a hexagon or triangle . This is the code my prof used but it makes a square, so I was wondering if I can adjust/modify the code to make a triangle or hexagon.

layer_1 = 255*ones(50, 50); % red layer; 100%turned on
layer_2 = 255*ones(50, 50); % green layer 100%turned on
layer_3 = 255*ones(50, 50); % blue layer 100%turned on
my_image(:,:,1) = layer_1; % red layer,applied to 3D matrix.
my_image(:,:,2) = layer_2; % green layer,applied to 3D matrix.
my_image(:,:,3) = layer_3; % blue layer,applied to 3D matrix.
imshow(my_image);
for row_index = 1:25
 for col_index = 25:50
 my_image(row_index,col_index,2) =0; % modify green layer values
 my_image(row_index,col_index,3) =0; % modify blue layer values
 end
end
imshow(my_image); 

[Image Analyst]

You saw the code in demo #2:

% Make a triangle on it.
triangleXCoordinates = [360 420 480];
triangleYCoordinates = [350 252 350];
traiangleBinaryImage = poly2mask(triangleXCoordinates, triangleYCoordinates, rows, columns);
% Burn it into the gray scale image.
grayImage(traiangleBinaryImage) = 255;

What's complicated about that?

And for the more general case with any number of vertices that you specify, demo #1 has this code:

%----------------------------------------------------------------------------------------------------------------------------------
% Create a single polygon with the specified number of sides in a binary image of the specified number of rows and columns.
% centroidToVertexDistance is the distance from the centroid to each vertex.
% If centroidToVertexDistance is a length 2 vector, then this indicated the minimum and maximum size range and
% it will create a random size polygon between the min and max distance.
function binaryImage = CreatePolygon(numSides, centroidToVertexDistance, rows, columns)
try
	% Get the range for the size from the center to the vertices.
	if length(centroidToVertexDistance) > 1
		% Random size between a min and max distance.
		minDistance = centroidToVertexDistance(1);
		maxDistance = centroidToVertexDistance(2);
	else
		% All the same size.
		minDistance = centroidToVertexDistance;
		maxDistance = centroidToVertexDistance;
	end
	thisDistance = (maxDistance - minDistance) * rand(1) + minDistance;
	
	% Create a polygon around the origin
	for v = 1 : numSides
		angle = v * 360 / numSides;
		x(v) = thisDistance * cosd(angle);
		y(v) = thisDistance * sind(angle);
	end
	% Make last point the same as the first
	x(end+1) = x(1);
	y(end+1) = y(1);
	% 	plot(x, y, 'b*-', 'LineWidth', 2);
	% 	grid on;
	% 	axis image;
	
	% Rotate the coordinates by a random angle between 0 and 360
	angleToRotate = 360 * rand(1);
	rotationMatrix = [cosd(angleToRotate), sind(angleToRotate); -sind(angleToRotate), cosd(angleToRotate)];
	% Do the actual rotation
	xy = [x', y']; % Make a numSides*2 matrix;
	xyRotated = xy * rotationMatrix; % A numSides*2 matrix times a 2*2 = a numSides*2 matrix.
	x = xyRotated(:, 1); % Extract out the x as a numSides*2 matrix.
	y = xyRotated(:, 2); % Extract out the y as a numSides*2 matrix.
	
	% Get a random center location between centroidToVertexDistance and (columns - centroidToVertexDistance).
	% This will ensure it's always in the image.
	xCenter = thisDistance + (columns - 2 * thisDistance) * rand(1);
	% Get a random center location between centroidToVertexDistance and (rows - centroidToVertexDistance).
	% This will ensure it's always in the image.
	yCenter = thisDistance + (rows - 2 * thisDistance) * rand(1);
	% Translate the image so that the center is at (xCenter, yCenter) rather than at (0,0).
	x = x + xCenter;
	y = y + yCenter;
	binaryImage = poly2mask(x, y, rows, columns);
catch ME
	errorMessage = sprintf('Error in function %s() at line %d.\n\nError Message:\n%s', ...
		ME.stack(1).name, ME.stack(1).line, ME.message);
	fprintf(1, '%s\n', errorMessage);
	uiwait(warndlg(errorMessage));
end

It's not that bad. Half the lines are comments. Most of my programs are thousands of lines long. Don't be scared of a function that's a dozen or two lines long. If you want, you could shorten it by taking out the code that rotates the shape by a random angle.

[Sakibur Rahman]

thanks I appreciate it, but what about using the code my teacher gave, we're supposed to use that but im not sure how.

[murat murat]

dear sir I run the code But there are some errors,like:

Error in ifunc parse_input[] at line 131

minpeak prominence is not a recognized parameter

and

Error in shape_recognition_demo1 at line 54

index exceeds matrix dimention

could you help me to solve the problem ?

[Image Analyst]

@murat murat attach your code. I need to see how you changed it because I just downloaded and ran the code I attahced and it ran fine. For example I had

[peakValues, peakIndexes] = findpeaks(distances, 'MinPeakProminence', minPeakHeight);

while you might have changed it to

[peakValues, peakIndexes] = findpeaks(distances, 'Min Peak Prominence', minPeakHeight);

[murat murat]

I did not do any changes, I only downloaded and run demo2 is working good but 1 is not .

I attached the errors and code

thank you very much

[Image Analyst]

@murat murat I just downloaded both and both ran fine. The only thing I can guess is that you have a really old version of MATLAB where the findpeaks() function does not have the 'MinPeakProminence' option. What version do you have? And if you look up the help for findpeaks() does it have the MinPeakProminence option?

[murat murat]

I am using matlab 2014 so that is the problem, :.(

thank you very much

[Image Analyst]

Then just try getting rid of that option and see if it still works.